Question

If the velocity of hydrogen molecule is $5\times {10}^{4}cmse{c}^{-1}$, then its de-Broglie wavelength is

• A. $2\mathring{A}$
• B. $4\mathring{A}$
• C. $8\mathring{A}$
• D. $100\mathring{A}$

Answer 1

The correct option is B

$4\mathring{A}$

According to de-Broglie

$\lambda =\frac{h}{mv}=\frac{6.62\times {10}^{-20}erg.sec}{\frac{2}{6.023\times {10}^{23}}\times 5\times {10}^{4}cm/sec}$

$=\frac{6.62\times {10}^{-27}\times 6.023\times {10}^{23}}{2\times 5\times {10}^4}cm=4\times {10}^{-8}cm=4\mathring{A}$.